DEPARTMENT OF MATHEMATICS

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1 Name(Last/First): GUID: DEPARTMENT OF MATHEMATICS Ma322005(Sathaye) - Final Exam Spring 2017 May 3, 2017 DO NOT TURN THIS PAGE UNTIL YOU ARE INSTRUCTED TO DO SO. Be sure to show all work and justify your answers. There are 6 problems and a total of 9 pages including the front page and two pages at the end for extra work. No other sheets, books, papers are allowed. Be sure to record any additional work on the last two pages saved for it. You may use backs of the main pages for scratch work, but this work will not be graded. If some work is saved on the last two pages, be sure to refer to it in the question on the main pages. Maximum Actual Problem Score Score Bonus 4 Total 100

2 Q.1: (2+2)+4+2+(3+3) 1. Let V and W be vector spaces defined over real numbers. Complete the following definitions and answer related questions. (a) A map L from V to W is defined to be a linear transformation if the following two conditions hold: Linearity. Scalar multiplication. Answer: Linearity: L(v 1 + v 2 ) = L(v 1 ) + L(v 2 ) and Scalar multiplication L(cv 1 ) = cl(v 1 ) for all v 1, v 2 V and scalars c. (b) Explicitly describe a map T : R 2 R 2 such that T is not a linear transformation. You must explain which condition fails and how. ( ) ( ) x x + 1 Answer: Many possible answers: Define L =. Both conditions fail when ( ) y y 1 we take v 1 = v 2 = and c = 2. 1 (c) Suppose that L : R n R d is defined by the formula L(X) = AX where A is a 6 3 matrix. Determine the values of n and d with explanation. Answer: For AX to be defined, we need X to have 3 entries, so n = 3. Similarly, for AX to be in R d A must have d rows, so d = 6. 1

3 (d) Write down a 3 3 matrix A with all non zero entries such that the linear transformation T A : R 3 R 3 defined by T A (X) = AX is not surjective (onto). You must explain why T A is not surjective. Answer: The dimension of the image of T A is the rank of A. So, we need a matrix A of rank less than 3. Take A of rank 1 given as Is it possible to construct the matrix A as described above such that the map T A is injective but not surjective? Either construct such a matrix A or explain why it cannot exist. Answer: This is not possible. The map T A is injective iff the three columns of A are independent iff rank(a) = 3. Similarly It is surjective exactly when the number 3 of rows equals the rank(a). Hence, injectivity holds iff surjectivity holds. 2

4 Q.2: Define a linear transformation L : P 3 P 3 by the formula L(p(t)) = tp (t) cp(t) where c is an unspecified scalar. Thus, for example, L(1 + t + t 2 + t 3 ) = (3 c) t 3 + (2 c) t 2 + (1 c) t c. Answer the following questions about this transformation. (a) Recall the standard basis B = ( 1 t t 2 t 3 ) for P 3. Calculate the matrix of L in the basis B. The matrix found shall be referred to as M in the following questions on this page. Answer: The images of the basis vectors (polynomials) are L(1) = t(0) c(1) = c, L(t) = t(1) c(t) = (1 c)t, L(t 2 ) = t(2t) ct 2 = (2 c)t 2 and L(t 3 = t(3t 2 ) c(t 3 ) = (3 c)t 3. The corresponding coordinate vectors give the matrix M =Answer: c c c c (b) Determine all the eigenvalues (which will be expressions involving c) for the matrix M. Moreover, for each eigenvalue, determine a corresponding eigenvector belonging to it. Answer: Since the matrix M λi is diagonal, the eigenvalues are given by the diagonal entries c, 1 c, 2 c, 3 c. The corresponding eigenvectors are the four columns of I 4 by the voodoo principle. 3

5 (c) Prove that L is diagonalizable (which is the same as M being diagonalizable.) Answer: First, since M is visibly diagonal, it is diagonalizable! You could also observe that R 4 has a basis of 4 eigenvectors of M, so L (or M) have eigendim 4. You could also use the fact that the four eigenvalues are distinct. (d) Determine a value of c, if possible, for which the matrix M has only three different eigenvalues. Justify your claim. Answer: This is not possible since the four eigenvalues are distinct for all values of c. c = Prove or disprove that L is still diagonalizable for the above chosen value of c. Answer: Yes, as shown above. 4

6 Q.3: 2+(2+2) You are given two matrices: P = 1 2 1, Q = Let v 1, v 2, v 3 be the three columns of P in order Use this information to answer the following questions. and told that P Q = I. (a) Assume that v 1 is an eigenvector for a 3 3 matrix A with eigenvalue 3. Calculate Av 1. The answer should be a vector in R 3. Answer: 3v 1 = Av 1 = (b) Also assume that v 2 is an eigenvector for the same A with eigenvalue 2 and v 3 similarly is an eigenvector for A with eigenvalue 1. Calculate Av 2, Av 3. Answer: Respectively, 2v 2 = Av 3 = , v 3 = Av 2 = (c) Determine the matrix A using the above information. Your final answer must be a single 3 3 matrix! You should first write A as a certain product of matrices and then the final answer is allowed to be from a calculator. Answer: Let D = gives A =Answer: Then AP = P D. Hence A = P DP 1 = P DQ. This. 5

7 (d) Complete this definition: A square matrix M is said to be diagonalizable if Answer: HMH 1 = D where H is some invertible matrix and D is a diagonal matrix. Prove that the above matrix A is diagonalizable by using your definition. Answer: Note that A = P DP 1 gives P 1 AP = D or QAP = D. 6

8 Q.4: Suppose that V is a vector space with a basis B = ( v 1 v 2 ) v 3. Further suppose that V has an inner product with the inner product matrix: < B, B >= Use this information to answer the following questions about vectors in V. (a) Determine the angle (in degrees) between the vectors v 1 and v 2 using the given inner product. Set up the formula before evaluation. Answer: The angle is given by ( θ = arccos < v 1, v 2 > < v1, v 1 >< v 2, v 2 > ). Evaluation gives: arccos( ) = (b) Set w 1 = v 1. Determine a scalar c such that w 1 and w 2 = v 2 + cv 1 are perpendicular under this inner product. Answer: 0 =< w 1, w 2 >=< v 1, v 2 > +c < v 1, v 1 >= 2 + c. So we get c =Answer: c = 2. (c) Determine the projection of v 3 onto the space spanned by w 1. Hint: The answer is a vector k 1 w 1 such that v 3 k 1 w 1 is perpendicular to w 1. Denote the projection vector by h 1. Answer: As above, we solve for k 1 in 0 =< w 1, v 3 k 1 w 1 >= 4 k 1 1. Thus k 1 = 4 and we get: h 1 = Answer: 4w 1. (d) Similarly determine the projection of v 3 onto the span of w 2 found above. Denote the new projection vector by h 2. Answer: Similarly we solve 0 =< w 2, v 3 k 2 w 2 > for k 2. The equation becomes 0 =< w 2, v 3 > k 2 < w 2, w 2 >= 4 k 2 (1). So k 2 = 4. h 2 = Answer: 4w 2. (e) Set w 3 = v 3 h 1 h 2. From your calculations, prove that w 3 is perpendicular to w 1 as well as w 2. Answer: By construction, v 3 h 1 is perpendicular to w 1 and h 2 = k 2 w 2 is clearly perpendicular to w 1, we see that w 3 is perpendicular to w 1. Similarly, observe that < v 3 h 2 as well as h 1 are perpendicular to w 2 and so w 3 is perpendicular to w 2. 7

9 Q.5: 2+(4+4) Suppose that W is a vector space with a basis C = ( w 1 w 2 w 3 ) w 4. Further suppose that V has an inner product with the inner product matrix: < C, C >= H = Work out the Gram-Schmidt algorithm as taught in the class notes by answering the following questions. (a) Write out the augmented matrix (H I 4 ). Double check your entries. Answer: (H I 4 ) = (b) Carry out the REF algorithm on the above matrix (H I 4 ) =. Only two steps are needed. Warning: Beware of calculator answers. They may not be compatible with the current algorithm. Follow the strict algorithm as taught. Answer:

10 (c) Using the REF, determine the matrix R which should be 4 4 unit upper triangular. Answer: This is obtained from transposing the matrix in the last four columns: R = Answer: (d) Report the new basis C = CR which should be a 4 element row of linear expressions in w 1, w 2, w 3, w 4. Answer: We get C = Answer: ( w1 w 1 + w 2 w 1 w 2 + w 3 w 4 ) (e) Finally determine the inner product matrix for the new basis C. Hint: You should have a diagonal matrix and no new calculations are needed! Answer: We simply pick up the diagonal of the final REF calculated above. < C, C >. Answer:

11 Q.6: Answer the following questions. In proving a result in some part, you may assume the above parts as proved. (a) Suppose that A is a 4 4 matrix with characteristic polynomial f(λ) = λ 4 5λ Determine all the eigenvalues of A. Hint: The polynomial is a quadratic polynomial in λ 2. Answer: First, it factors as a polynomial in λ 2 : (λ 2 2)(λ 2 3). Further factorization of the quadratics gives 4 distinct eigenvalues: 2, 2, 3, 3. Eigenvalues are: (b) Prove that A is diagonalizable. Explain your argument. Answer: Since we have 4 = Colnum(A) distinct eigenvalues, A is diagonalizable. (c) Any square matrix P which satisfies the equation P 2 P = 0 is said to be a projection matrix. Suppose that P is a projection matrix. If v = P w for some vector w, then prove that P v = v. Hint: A matrix A is a zero matrix if and only if Aw = 0 for all w (of compatible size). Answer: Since v = P w the needed equation becomes P 2 w = P w or (P 2 P )w = 0. This is true since P 2 P is zero. (d) Prove that 1 and 0 are the only possible eigenvalues for a projection matrix P. Hint: Suppose that u is an eigenvector belonging to an eigenvalue c. Use the given condition to evaluate (P 2 P )u. Answer: By the hint, we get 0 = (P 2 P )u = (c 2 c)u. Since u 0, we have c 2 = c, i.e. c = 0 or c = 1. (e) Suppose that P is a non zero projection matrix. Use the above result to show that every non zero vector in Col(P ) is an eigenvector for eigenvalue 1. Answer: A vector v is in the Col(P ) iff v = P X for some X. Then as calculated above, P v = P 2 X = P X = v. Hence every non zero v is an eigenvector belonging to eigenvalue 1. (f) Extra credit 8 points: This extra credit work is best written on page 8. Consider the linear transformation T P : R n R n where P is a projection matrix. Prove that Ker(T P ) = Nul(P ) and Image(T P ) = Col(P ). Answer: This is immediate by definitions. Ker(T P ) is the set of all vectors v such that T P (v) = P v = 0, so it is the space Nul(P ). Similarly Image(T P ) is the set of all vectors w of the form w = T P (v) = P v. So clearly, this is exactly the space Col(P ). Also prove that I P is a projection matrix as well. How are its Kernel and Image related to those of P? Answer: Set Q = I P and calculate Q 2 Q. This gives (I P )(I P ) (I P ) which simplifies to I P P + P 2 I + P = I P P + P I + P = 0 as needed. Prove or disprove Every projection matrix is diagonalizable! Answer: Let us determine the eigenspaces of each of the possible eigenvalues 0 and 1. As shown above, The eigenspace of 1 is simply the Col(P ). So, its dimension is rank(p ). Similarly, the eigenspace of 0 is the Nul(P ) and its dimension is Colnum(P ) rank(p ). The sum of these dimensions is Colnum(P ) so P is diagonalizable. 10

12 Extra Page 1 Name(Last/First): GUID: This page must not be removed! This page is to be used to submit work which did not fit on the problem pages 1 to 4. Important: You must have clear reference to work displayed here. Without that, this page will be ignored. 11

13 Extra Page 2 Name(Last/First): GUID: This page must not be removed! This page is to be used to submit work which did not fit on the problem pages 1 to 4. Important: You must have clear reference to work displayed here. Without that, this page will be ignored. 12